Menu Close

calculate-n-1-n-n-1-2-n-2-

Tinku Tara 0 Comments
Pin




Question Number 40897 by abdo.msup.com last updated on 28/Jul/18
calculate Σ_(n=1) ^∞ (n/((n+1)^2 (n+2)))
calculate∑n=1∞n(n+1)2(n+2)
Commented by math khazana by abdo last updated on 03/Aug/18
let decompose F(x)=(x/((x+2)(x+1)^2 )) F(x)=(a/(x+2)) +(b/(x+1)) +(c/((x+1)^2 )) a=lim_(x→−2) (x+2)F(x)= −2 c=lim_(x→−1) (x+1)^2 F(x)=−1 ⇒ F(x)=((−2)/(x+2)) +(b/(x+1)) −(1/((x+1)^2 )) F(0)=0=−1 +b −1=b−2⇒b=2 ⇒ F(x)=((−2)/(x+2)) +(2/(x+1)) −(1/((x+1)^2 )) S_n =Σ_(k=1) ^n (k/((k+2)(k+1)^2 )) =Σ_(k=1) ^n F(k) =−2Σ_(k=1) ^n (1/(k+2)) +2Σ_(k=1) ^n (1/(k+1)) −Σ_(k=1) ^n (1/((k+1)^2 )) but Σ_(k=1) ^n (1/(k+2)) = Σ_(k=3) ^(n+2) (1/k)=H_(n+2) −(3/2) Σ_(k=1) ^n (1/(k+1)) =Σ_(k=2) ^(n+1) (1/k) =H_(n+1) −1 Σ_(k=1) ^n (1/((k+1)^2 )) =Σ_(k=2) ^(n+1) (1/k^2 ) =ξ_(n+1) (2)−1 ⇒ S_n =−2{H_(n+2) −(3/2)} +2{H_(n+1) −1} −ξ_(n+1) (2) +1 =−2(H_(n+2) −H_(n+1) ) +2 −ξ_(n+1) (2) but lim_(n→+∞) H_(n+2) −H_(n+1) =0 and lim_(n→+∞) ξ_(n+1) (2)=(π^2 /6) ⇒ S=lim_(n→+∞) S_n =2−(π^2 /6) .
letdecomposeF(x)=x(x+2)(x+1)2F(x)=ax+2+bx+1+c(x+1)2a=limx→−2(x+2)F(x)=−2c=limx→−1(x+1)2F(x)=−1⇒F(x)=−2x+2+bx+1−1(x+1)2F(0)=0=−1+b−1=b−2⇒b=2⇒F(x)=−2x+2+2x+1−1(x+1)2Sn=∑k=1nk(k+2)(k+1)2=∑k=1nF(k)=−2∑k=1n1k+2+2∑k=1n1k+1−∑k=1n1(k+1)2but∑k=1n1k+2=∑k=3n+21k=Hn+2−32∑k=1n1k+1=∑k=2n+11k=Hn+1−1∑k=1n1(k+1)2=∑k=2n+11k2=ξn+1(2)−1⇒Sn=−2{Hn+2−32}+2{Hn+1−1}−ξn+1(2)+1=−2(Hn+2−Hn+1)+2−ξn+1(2)butlimn→+∞Hn+2−Hn+1=0andlimn→+∞ξn+1(2)=π26⇒S=limn→+∞Sn=2−π26.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *